CF Round 4 div2B. Before an Exam(DP, 背包)

回复手感的第一天

怎么正解都是贪心啊，这题明明dp可以做（

首先题目问给定若干个区间，每个区间选择一个数字，要求构造一种情况使得和为k，并且给出方案，否则输出NO

看了下数据不是很大，似乎可以用背包做，然后记录转移路径就好

然后发现内存超了，我索性把map当作数组用，结果水过去了2333

#include <bits/stdc++.h>
using namespace std;
#define limit (200000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a  ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
typedef long long ll;
typedef unsigned long long ull;
char buf[1<<23],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf;
inline ll read(){
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
    ll sign = 1, x = 0;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
    return x * sign;
#undef getchar
}//快读
void print(ll x) {
    if(x/ 10) print(x / 10);
    *O++=x % 10+'0';
}
void write(ll x, char c = 't') {
    if(x < 0)putchar('-'),x = -x;
    print(x);
    if(!isalpha(c))*O++ = c;
    fwrite(obuf,O-obuf,1,stdout);
    O = obuf;
}
int kase;
int n, m,k;
int a[limit];
int l[limit], r[limit];
map<int, map<int, int>> dp, inc;
int tot;
void solve(){
    cin>>n>>k;
    rep(i,1,n){
        cin>>l[i]>>r[i];
    }
    rep(i,l[1],r[1]){
        dp[1][i] = 1;
        inc[1][i] = i;
    }
    rep(i,2,n){
        rep(j,l[i],r[i]){
            per(h, j, k){
                if(dp[i][h])continue;
                dp[i][h] = dp[i - 1][h - j];

                if(dp[i][h]){
                    inc[i][h] = j;
                }
            }
        }
    }
    if(!dp[n][k]){
        cout<<"NO"<<endl;
    }else{
        int target = k;

        deque<int>ans;
        per(i,1,n){
//            cout<<dp[i][target]<<" "<<inc[i][target]<<" "<<target<<endl;
            ans.push_front(inc[i][target]);
//            cout<<i<<" "<<target<<endl;
            target -= inc[i][target];


        }
        cout<<"YES"<<endl;
        for(auto it : ans){
            cout<<it<<" ";
        }
        cout<<endl;
    }



}
int32_t main() {
#ifdef LOCAL
    FOPEN;
//    FOUT;
#endif
    FASTIO
//    cin>>kase;
//    while (kase--)
    solve();
    cerr << "Time elapsed: " << 1.0*clock()/CLOCKS_PER_SEC << "s\n";
    return 0;
}